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I made a mistake with the $\frac{18}{xyz}$, though I am scared to edit the latex : Do you think you can manipulate $x^2 y^2 + y^2 z^2 + x^2 z^2$ into something you already know? The answer follows from there.

$(x+y+z)^2 - 2(xy + xz + yz) = x^2 + y^2 + z^2$

$(x+y+z)^2 - 2(-7) = 18$

$(x+y+z)^2 = 4$

$x+y+z = 2$

$x^2 y^2 + 2 x^2 y z + x^2 z^2 + 2 x y^2 z + 2 x y z^2 + y^2 z^2 = 49$

$x^2 y^2 + x^2 z^2 + y^2 z^2 + 2xyz(x + y + z) = 49$

$\frac{xyz^2}{z^2} + \frac{xyz^2}{y^2} + \frac{xyz^2}{x^2} + 4xyz = 49$

$xyz \left(\frac{xyz}{z^2} + \frac{xyz}{y^2} + \frac{xyz}{x^2} \right) + 4xyz = 49$

$xyz \left(\frac{xyz(x^2 y^2 + y^2 z^2 +x^2 z^2)}{xyz^2} \right) + 4xyz = 49$

$xyz \left(\frac{18}{xyz} \right) + 4xyz = 49$

$18 + 4xyz = 49$

$4xyz = 31$

$xyz = \boxed{\frac{31}{4}}$

Using the  sum of differences

1     9       36     100     225    441     784   

   8     27       64      125     216    343

      19      37       61     91      127

           18      24      30      36

                 6        6        6

We have  4  non-zero  differences

The polynomial  will be of  the  form

an^4  + bn^3  + cn^2  + dn  +  e  

We  have  this system

a  +  b  +  c  +   d  +  e    =  1

16a + 8b + 4c + 2d + e  = 9

81a + 27b+ 9c + 3d + e  =  36

256a + 64b + 16c + 4d  + e  = 100

625a + 125b + 25c + 5d  + e  = 225

This is  tedious to solve....using a little  technology

a = 1/4

b = 1/2

c = 1/4

d,e  =  0

The polynomial is

f(n)  =  (1/4)n^4  + (1/2)n^3  + (1/4)n^2

Minimum: You want the least numbers possible, and for those numbers you want the least of your set, namely 0 and 1. 

Maximum: You want the most numbers possible, and there isn't an upper bound, so 0 + 1 + 2 + 3 + 4 + 5 = 15 is your max.

Thus, [1, 15]

Verify:

0+1

0+2

0+3

0+4

0+5

1+5

2+5

3+5

4+5

1+2+3+4

1+2+3+5

1+2+4+5

1+3+4+5

2+3+4+5

1+2+3+4+5

Review your class transcript. This should give you a hint on how to start. Also look up the definitions and formulas relating to terminal points, and then with those you can find your answer.

$\lceil{\sqrt{300}}\rceil = \lceil{\sqrt{3 \cdot 100}}\rceil = \lceil{10 \sqrt{3}}\rceil \approx \lceil{10 \cdot 1.73}\rceil = \lceil{17.3}\rceil = \boxed{18}$

We can express any integer  sum from  1  to  15  inclusive

6A + 10P  =  14A  +  5P

10P  - 5P  =  14A  - 6A

5P  =  8A    

P  = (8/5)A

So

6A  +  10P   =

6A  +  10  ( 8/5)A

6A  +  (80/ 5)A

6A  + 16A  =

22A   ⇒    she could have  bought 22 apples

That was quick, thank you so much!

Think of the perfect squares around 300. 17^2 = 289, 18^2 = 324. \(\sqrt{324} = 18\) as we know, so 18 is the least integer greater than $\sqrt{300}$

Let  x  be  the number of   10 cent  coins   and   88 - x  the  number of 1 dollar  coins

So

.10x =   1 ( 88 - x)

.10x  =  88- x

x + .10x = 88

1.1 x  = 88

88 / 1.1   = x  =  80       (10 cent coins)

This means y = 100

\(-6t^2+51 = 100\)

\(-6t^2+51 =100 = 0\)

Apply quadratic formula to get the positive solution of \(\frac{17}{4}+\frac{\sqrt{201}}{12}\)

Thank you so much!

g(16)  =  16^2   =   256

f ( g (16))  =   f (256)  =  3 - x^2  =     3 - 256^2   =     -65533

Plugging in the values,

\(\sqrt{7} \cdot \sqrt{13} \cdot \sqrt{91} \cdot \sqrt{97}\)

\(\sqrt{91} \cdot \sqrt{91} \cdot \sqrt{97}\)

So we get \(91\sqrt{97}\) as the answer

W     10         X

                            18

          Z                          Y

1/2   of 90  =  45°

Because we  have  a parallelogram, angle  Y =  angle W     and  YZ = 10

So

Area  =  XY * YZ  * sin (45°)  =   (18) (10)  sqrt (2)   / 2       =   90*sqrt (2)

AB^2  +   BC^2   =  AC^2

5^2   +  8^2    =   89 = AC^2   =   area of the square

Let's put this into equations,

\(m + 4t = 13\)

\(m + 7t = 19\)

Subtracting the equations we get,

\(3t = 6\)

\(t = 2\)

Now we know the cost of a towel, we can plug it back into one of the equations.

\(m + 8 = 13\)

\(m = 5\)

So a mug costs 5 dollars.

Melody gave a great explanation for a problem very similar. Follow their solution for your values: https://web2.0calc.com/questions/binomial-probability_9

I think there may be a diagram or picture because I don't know what T is. Attach that please

First let's take the prime factorization of 1584,

\(2^4\cdot 3^2\cdot11 = 1584\)

To make \(1584\cdot x \), each of the exponents in the prime factoriziation of the cube need to be a multiple of 3. To make that, the prime factoriziation of x needs to create that by having the correct exponents of,

\(2^2 \cdot 3 \cdot 11^2 = 1452\)

Making the cube's prime factorization,

\(2^6 \cdot 3^3 \cdot 11^3 = 2299968\)

\(\sqrt[3]{2299968} = 132\)

So our cube is actually a cube. Now that we know that x = 1452, we can use LCM to find a common multiple of 1452 and 1584 to find y. 

The LCM of 1452 and 1584 is 17424.

\(\frac{17424}{1452} = 12\)

So y = 12. 

Sorry if my method was kind of confusing lol

The logical starting point  is  to look  at  the  4  digit perfect squares where  the  square root is a multiple of 3

We find ( through a little trial and error)

54^2   =  2916 =  n

2 + 9  + 1 +  6 =    18

18*3    = 54

Just a quadratic, so we have,

\(-4.9t^2 - 3.5t + 12.4 = 0\)

Apply quadratic formula,

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {3.5 \pm \sqrt{255.29} \over -9.8}\)

Simplify via calculator we get roots, \(x = -1.99, 1.28\). We want the positive root so the answer is 1.28

ummm answer is 7 but thanks for giving me a starting point!